How many grams of magnesium metal will react completely with 6.3 liters of 4.5 M HCl? Show all of the work needed to solve this problem.

Mg (s) + 2 HCl (aq) yields MgCl2 (aq) + H2 (g)

Question

How many grams of magnesium metal will react completely with 6.3 liters of 4.5 M HCl? Show all of the work needed to solve this problem. 

Mg (s) + 2 HCl (aq)  yields MgCl2 (aq) + H2 (g)

abmon98 · Answer

$$Mg(s)+2HCL(aq)\rightarrow MgCl2+H2(g) $$
Number of moles=Volume(l)*Concentration(mol/l)
Number of moles of HCL=6.3*4.5=28.35 moles of HCl
Look at the mole ratio according to the balanced chemical equation. 1 Mg mole reacts with 2 mole of HCl to produce 1 mole of both H2 and MgCl2
1:2:1:1
now lets in account the mole ratio between the two reactants only.
1:2
x:28.35
x=14.175 
Use Number of moles=Mass(g)/Molar Mass(g/mol)
Atomic weight of Mg:24

anonymous · Answer

Mg + 2 HCl → H2 + MgCl2
(6.3 L) x (4.5 mol/L HCl) x (1 mol Mg / 2 mol HCl) x (24.30506 g Mg/mol) =
345 g Mg